Theorem sbc3or 42105

Description: sbcor 3772 with a 3-disjuncts. This proof is sbc3orgVD 42424 automatically translated and minimized. (Contributed by Alan Sare, 31-Dec-2011.) (Revised by NM, 24-Aug-2018.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
sbc3or	⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒))

Proof of Theorem sbc3or

Step	Hyp	Ref	Expression
1		sbcor 3772	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ∨ [𝐴 / 𝑥]𝜒))
2		df-3or 1086	. . . . 5 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
3	2	bicomi 223	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ (𝜑 ∨ 𝜓 ∨ 𝜒))
4	3	sbcbii 3780	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ [𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒))
5		sbcor 3772	. . . 4 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓))
6	5	orbi1i 910	. . 3 ⊢ (([𝐴 / 𝑥](𝜑 ∨ 𝜓) ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
7	1, 4, 6	3bitr3i 300	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
8		df-3or 1086	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
9	7, 8	bitr4i 277	1 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒))

Syntax hints:   ↔ wb 205   ∨ wo 843   ∨ w3o 1084  [wsbc 3719

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-10 2140  ax-12 2174  ax-ext 2710

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3or 1086  df-tru 1544  df-ex 1786  df-nf 1790  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-v 3432  df-sbc 3720

This theorem is referenced by:  sbcoreleleq  42108